1) Point charges q, a) Find the electric field at P. (b) Find the magnitude of the electric field at P. c) Find the direction of the electric field at P. d) Find the potential at P. (e) Find the electrical force on the charge q.. (k = 9x10° N×m2/C², and n= 10-9) 100 nC, q, = -45 nC, and q, = 32 nC are arranged in Figure. %3D |3D %3D

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The questions are in the pictures I couldn't help please

Bul
1) Point charges q,
(a) Find the electric field at P.
(b) Find the magnitude of the electric field at P.
(c) Find the direction of the electric field at P.
(d) Find the potential at P.
(e) Find the electrical force on the charge q1. (k = 9×10° N×m²/C², and n = 10-9)
100 nC, q, = -45 nC, and q, = 32 nC are arranged in Figure.
%3D
100 nC
-45 nC
92
3 m
3 m
4 m
P.
43 = 32 nC
2304
&610
28.04.2021
Transcribed Image Text:Bul 1) Point charges q, (a) Find the electric field at P. (b) Find the magnitude of the electric field at P. (c) Find the direction of the electric field at P. (d) Find the potential at P. (e) Find the electrical force on the charge q1. (k = 9×10° N×m²/C², and n = 10-9) 100 nC, q, = -45 nC, and q, = 32 nC are arranged in Figure. %3D 100 nC -45 nC 92 3 m 3 m 4 m P. 43 = 32 nC 2304 &610 28.04.2021
Bul
2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric
constants K1
2 and K2
4, as shown in Figure. Parallel-plates have the area A = 0.8 m². The
distance is d=0.04 m. A potential difference of AV = 24 V is applied to the circuit.
(a) Find the equivalent capacitance of the system.
(b) Find the potential AV,. (E,-8.85×10-12 C²/N×m?)
K1
A/2
AV
K2
A/2
K2
A/2
2d/3 d/3
K2
A/2
d.
AV = 24 V
23:04
AG 0
28.04.2021
----->
Transcribed Image Text:Bul 2) Two parallel-plate capacitors are constructed using dielectric materials with dielectric constants K1 2 and K2 4, as shown in Figure. Parallel-plates have the area A = 0.8 m². The distance is d=0.04 m. A potential difference of AV = 24 V is applied to the circuit. (a) Find the equivalent capacitance of the system. (b) Find the potential AV,. (E,-8.85×10-12 C²/N×m?) K1 A/2 AV K2 A/2 K2 A/2 2d/3 d/3 K2 A/2 d. AV = 24 V 23:04 AG 0 28.04.2021 ----->
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